# 352 - Topic 06 - First order linear DEs

## 3299 days ago by Professor352

John Travis

Mississippi College

Playing around with solving first order linear Differential Equations.

We will presume the DEs are in the form $\frac{dy}{dt} = P(t) y + Q(t)$

reset()
# Solving a first order linear DE in general using Sage var('t') y = function('y',t) P = function('P',t) Q = function('Q',t) DE = diff(y,t)==P*y+Q soln=desolve(DE,y,ivar=t) show(soln)
 \newcommand{\Bold}[1]{\mathbf{#1}}{\left(c + \int e^{\left(-\int P\left(t\right)\,{d t}\right)} Q\left(t\right)\,{d t}\right)} e^{\left(\int P\left(t\right)\,{d t}\right)} \newcommand{\Bold}[1]{\mathbf{#1}}{\left(c + \int e^{\left(-\int P\left(t\right)\,{d t}\right)} Q\left(t\right)\,{d t}\right)} e^{\left(\int P\left(t\right)\,{d t}\right)}
# Now, let's try specifying values for P(t) and Q(t) and see what happens var('t,c') t0 = 0 # starting point for time t1 = 3 # ending point for time y0 = 1 # starting y-value y = function('y',t) P = function('P',t) Q = function('Q',t) P = -1/2 Q = 2*cos(t) DE = diff(y,t)==P*y+Q soln=desolve(DE,y,ivar=t,ics=[t0,y0]) show(soln.expand())
G=points(desolve_rk4(DE, y, ics=[t0,y0], ivar=t, end_points=3, step=0.25)) G+=plot(soln,t,[t0,t1]) show(G)
show(integrate(2*cos(t)*e^(-1/2*t),t).expand())
var('w') Plot1=plot_slope_field(1/2*w+2*cos(t),(t,0,10),(w,-6,6))+plot(soln,t,0,3,color='red',thickness=5)+G Plot1.show()